Since the series converges, an converges to zero. If nan doesn't converge to 0, then that means that an converges to zero no faster than 1/n. But if an converges no faster than 1/n, then the series will be no "better" than than the harmonic series, and so it certainly will not converge - contradiction. This is the rough idea you might want to work with, but of course, it's not rigorous yet.

The ratio test is for the series na_n, not the sequence na_n. The fact is that there exists converging a_n such that [tex]\sum a_n[/tex] converges but [tex]\sum na_n[/tex] does not. 1/n^2, for instance.

CASE 1:
Suppose {nan}n in N converges to some finite c > 0. Suppose that for all N > 0, there exists n > N such that nan < c/2. Well then {nan}n in N could not converge to c, countradicting our assumption. So there is some N > 0 such that for all n > N, nan> c/2. Thus we can choose a subsequence {nan}n > N which has each entry is at least c/2.

CASE 1:
Suppose {nan}n in N converges to some finite c > 0. Suppose that for all N > 0, there exists n > N such that nan < c/2. Well then {nan}n in N could not converge to c, countradicting our assumption. So there is some N > 0 such that for all n > N, nan> c/2.

If it converges to any number, then I can prove that number must be 0:

However, [tex]na_n[/tex] need not converge! All we know is that it's bounded, and therefore by Bolzano-Weierstrass theorem, there exists a converging subsequence. However, the above proof, using the converging subsequence, fails, since [tex]\sum\frac{c}{n_k}[/tex] might be convergent, and thus not leading into a contradiction.

The problem with your proof is that I can't see how you can conclude for all n>N, since it should be a subsequence only.

Then there is some [itex]\epsilon > 0 [/itex] such that for any [itex]n \in \mathbb{N}[/itex] there is some [itex]N > n[/itex] such that [tex]Na_N > \epsilon[/tex].

So, clearly there is an [itex]n_1>3[/itex] so that [itex]n_1a_{n_1}>\epsilon[/itex], and for any [itex]i>1[/itex] there is some [itex]n_i[/itex] such that [itex]n_i>2\times n_i-1[/itex] and [itex]n_ia_{n_i}>\epsilon[/itex].

Now [itex]\sum_{j=\ciel{\frac{n_i}{2}}}^{n_i} a_n > \frac{\epsilon}{4}[/itex]
since the sum contains at least [itex]\frac{n_i}{4}[/itex] terms all greater than [itex]\frac{\epsilon}{n_i}[/itex].

Since [itex]n_i<\frac{n_{i+1}}^{2}[/itex] we have [itex]n_i<\ceil(\frac{n_{i+1}}{2})[/itex]

What do you mean "all we know is that it's bounded." A priori, it might be unbounded, as it could be that an = 1. On the other hand, since we're asked to prove that {nan} converges to 0, we "know" that not only is it bounded, but that it converges to 0.

The problem with your proof is that I can't see how you can conclude for all n>N, since it should be a subsequence only.

No, I started with:

Suppose that for all N > 0, there exists n > N such that nan < c/2

which is:

[tex](\forall N > 0)(\exists n > N)(na_n < c/2)[/tex]

I derived a contradiction from this, so I may infer its negation, which is:

CASE 1:
Suppose {nan}n in N converges to some finite c > 0. Suppose that for all N > 0, there exists n > N such that nan < c/2. Well then {nan}n in N could not converge to c, countradicting our assumption. So there is some N > 0 such that for all n > N, nan> c/2. Thus we can choose a subsequence {nan}n > N which has each entry is at least c/2.
...
CASE 3:
{nan}n in N diverges. Same argument as in Case 1. Contradiction.

I don't see how the argument from case 1 applies to case 3 since there is, by definition, no well-defined limit for a divergent sequence. Moreover, there are divergent sequences where there is no suitable [itex]c[/itex].

For example, consider that if [itex]a_n=\frac{||-1^n+1|+.5^n}{n}[/itex], then [itex]\left{na_n\right}[/itex] is divergent since it has subsequences that converge to 0 and 2 (the odd and even terms respectively). For any finite [itex]c>0[/itex] and any [itex]n \in \mathbb{N}[/itex], there is some [itex]n>N[/itex] so that [itex]na_n<\frac{c}{2}[/itex] since the odd terms converge to zero.

Sorry, case 1 doesn't exactly apply to case 3. It does to the subcase that the sequence diverges to infinity. In this case, pick any finite c > 0, and the same argument applies, but you'd replace "could not converge to c" in the second sentence with "could not diverge to infinity". However, case 1 doesn't apply to the subcase where the sequence neither converges nor diverges to infinity, but is simply bounded divergent (i.e., it "oscillates"). Note that Treadstone proved that the sequence is bounded, so we don't really need to apply case 1 to the first subcase of case 3 (the subcase where the sequence diverges to infinity), we can just say that this subcase will never happen

since {an} converges, which is true since the series converges. At least, I think that's correct. If it is, then I think it can be used to argue that {nan} in fact does not oscilate, thereby eliminating the remaining subcase of case 3.

The [itex]\leq[/itex] are justified because [itex]a_n[/itex] is strictly positive, [itex]a_n[/itex] is monotone decreasing, and [itex]a_n[/itex] is strictly positive respectively.
Now we have
[tex] 0 < na_n \leq \sum_{i=\lfloor\frac{n}{2} \rfloor}^{\infty}{a_i}[/tex]

Since the sequence converges, the limit on the R.H.S. as n goes to infinity must be zero. This bounds [itex]na_n[/itex] above and below with sequences that go to zero, therfore [itex]na_n[/itex] goes to zero.